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(*CacheID: 232*)


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(*NotebookOptionsPosition[     48178,       1368]*)
(*NotebookOutlinePosition[     48812,       1391]*)
(*  CellTagsIndexPosition[     48768,       1387]*)
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Notebook[{

Cell[CellGroupData[{
Cell["General Routines", "Section"],

Cell[BoxData[
    \(discretize::usage = \n\t
      "\<Returns a discretization of the interval [a,b] into nCells cells.  \
The number of nodes returned is nCells+1.\>"; \n\n
    discretize[a_, b_, nCells_] := 
      Module[{\[CapitalDelta]x}, \n\t\t
        \[CapitalDelta]x = \((b - a)\)\/nCells; \n\t\t
        Return[Table[N[a + \[CapitalDelta]x\ i], {i, 0, nCells}]]; \n\t\t]; 
    \)], "Input"],

Cell[BoxData[
    \(initialCondition::usage = 
      "\<Returns a set of points {xi,yi} that represent the discretized \
initial conditions of a 1D finite difference problem.\>"; \n\n
    initialCondition[a_, b_, nCells_, iCFunction_] := 
      Module[{xIC, yIC}, \n\t\txIC = discretize[a, b, nCells]; \n\t\t
        yIC = Map[iCFunction, xIC]; \n\t\tReturn[Thread[{xIC, yIC}]]; \n
        \t\t]; \)], "Input"],

Cell["\<\
To obtain the initial conditions we start the scheme with the \
solution vector at t_0 and t_{-1} the same i.e. both icy.\
\>", "Text"],

Cell[BoxData[
    \(\(advectionOneDimension[xL_, \ xR_, \ iCFunction_, \ nCells_, \ 
        scheme_, \ \[Sigma]_, nTimeSteps_]\  := 
      Module[{ic, \ icx, \ icy}, \n\t\t
        ic\  = \ initialCondition[xL, \ xR, \ nCells, \ iCFunction]; \n\t\t
        icx\  = \ Map[First, \ ic]; \n\t\ticy\  = \ Map[Last, \ ic]; \n
        \t\t\t\t\t
        yAtnTimeSteps = 
          FixedPoint[scheme[#, \[Sigma]]&, {icy, icy}, nTimeSteps]; \n\t\t\ 
        Return[Thread[{icx, \ yAtnTimeSteps[\([1]\)]}]]; \n\t\t]; \)\)], 
  "Input"]
}, Open  ]],

Cell[CellGroupData[{

Cell["Page 339", "Section"],

Cell[CellGroupData[{

Cell["Problem 8.6-8.8", "Subsection"],

Cell["Exact solution:", "Text"],

Cell[BoxData[
    \(\(uExact[a_, \ k_, \ t_]\  := \n\ \ \ \ 
      Plot[\n\t\t
        If[0 < x - a\ t < 1, \n\t\tSin[k\ \[Pi]\ \((x\  - \ a\ t)\)], \n\t\t\t
          0], \ {x, \ \(-0.5\), \ 2.5}, \n\ \ \ \ \ \ 
        PlotStyle\  -> \ RGBColor[0, \ 1, \ 0], \n\ \ \ \ \ \ 
        DisplayFunction\  -> \ Identity]; \)\)], "Input"],

Cell[BoxData[{
    \(plotTimes = Table[1\/10\ i, \ {i, \ 0, \ 10}]; \ \ 
     (*{0, \ 0.1, \ 1, \ 2}; *) \ \n\), 
    \(temp\  = 
      Table[uExact[1, \ 1, \ plotTimes[\([i]\)]], 
        \ {i, \ 1, Length[plotTimes]}]; \n
    Show[temp, \ DisplayFunction\  -> \ $DisplayFunction]; \)}], "Input"],

Cell["\<\
Here we investigate how to get an second condition needed for the \
leapfrog method.  I tried to invert the difference scheme.  This seamed to \
work but I think that many terms are needed to have this transformation work.\
\
\>", "Text"],

Cell[BoxData[
    \(ic = initialCondition[0, 1, 50, Sin[\[Pi]\ #]&]; \nicy = Map[Last, ic]; 
    \n\nf1[v_List, \[Sigma]_] := 
      1\/\((1 + \[Sigma])\) - 
        \(\[Sigma]\/\((1 + \[Sigma])\)\^2\) RotateRight[v] + 
        \(\[Sigma]\^2\/\((1 + \[Sigma])\)\^3\) RotateRight[v, 2]; \n
    f2[v_List, \[Sigma]_] := 
      1 - \[Sigma] + \[Sigma]\^2 + 
        \((\(-\[Sigma]\) + 2\ \[Sigma]\^2)\) RotateRight[v] + 
        \(\[Sigma]\^2\) RotateRight[v, 2]; \n\nlp0 = ListPlot[icy]; \n
    lp1 = ListPlot[f1[icy, 0.5]]; \nlp2 = ListPlot[f2[icy, 0.5]]; \n
    Show[{lp0, lp2}]; \)], "Input"],

Cell["\<\
Another option would be just to use the same vector for the other \
condition needed for the leap frog method.\
\>", "Text"],

Cell["Numerical schemes:", "Text"],

Cell[BoxData[
    \(\(upwind[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{}, \n\t\t
        \(Return[\n
          \t\t\t{\((1 - \[Sigma])\)\ v1 + \[Sigma]\ RotateRight[v1], v1}\n
          \t\t\t\t]; \)\n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(laxFriedrichs[{v1_List, v0_List}, \ \[Sigma]_]\  := 
      Module[{}, \n\t\t
        \(Return[\n
          \t\t\t{\(1\/2\) 
              \((\((1\  + \ \[Sigma])\)\ RotateRight[v1]\  + \ 
                  \((1\  - \ \[Sigma])\)\ RotateLeft[v1])\), v1}\n\t\t\t]; 
        \)\n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(laxWendroff[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{}, \n\t\t
        \(Return[\n
          \t\t\t{\(\[Sigma]\/2\) \((1 + \[Sigma])\) RotateRight[v1] + 
              \((1 - \[Sigma]\^2)\) v1 + 
              \(\[Sigma]\/2\) \((\(-1\) + \[Sigma])\) RotateLeft[v1], v1}\n
          \t\t]; \)\n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(leapFrog[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{}, \n\t\t
        \(Return[\n
          \t\t\t{v0 - \[Sigma]\ RotateLeft[v1] + \[Sigma]\ RotateRight[v1], 
            v1}\n\t\t]; \)\n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(Exercise[xL_, xR_, k_, scheme_, \ nCells_, \ \[Sigma]_, \ nTimeSteps_]
         := \ Module[{a\  = \ 1, \ 
          \[CapitalDelta]x\  = \ \((xR\  - \ xL)\)\/nCells, \ ue, \ au, \ 
          lpa}, \ \[IndentingNewLine]ue\  = \ 
          uExact[a, \ k, \[Sigma] \( \[CapitalDelta]x\/a\) nTimeSteps]; \ 
        \[IndentingNewLine]ua\  = \ 
          advectionOneDimension[xL, \ xR, 
            Function[x, If[0 < x < 1, Sin[k\ \[Pi]\ x], 0]], \ nCells, \ 
            scheme, \ \[Sigma], \ nTimeSteps]; \ 
        \[IndentingNewLine]lpa\  = \ 
          ListPlot[ua, DisplayFunction\  -> \ Identity]; \n\t\t\n\t\t
        Show[{ue, \ lpa}, \ \n\ \ \ \ \ \ \ \ \ \ 
          DisplayFunction\  -> \ $DisplayFunction\n\t\t\t]; \n\t\t]; \)\)], 
  "Input"],

Cell[BoxData[{
    \(Exercise[\(-.5\), 2.5, 1, laxFriedrichs, 150, 0.25, 10]\), 
    \(Exercise[\(-.5\), 2.5, 10, laxFriedrichs, 150, 0.25, 10]\)}], "Input"],

Cell[BoxData[{
    \(Exercise[\(-.5\), 2.5, 1, upwind, 150, 0.25, 10]\), 
    \(Exercise[\(-.5\), 2.5, 10, upwind, 150, 0.25, 10]\)}], "Input"],

Cell[BoxData[{
    \(Exercise[\(-.5\), 2.5, 1, leapFrog, 150, 0.25, 10]\), 
    \(Exercise[\(-.5\), 2.5, 10, leapFrog, 150, 0.25, 10]\)}], "Input"],

Cell[BoxData[{
    \(Exercise[\(-.5\), 2.5, 1, laxWendroff, 150, 0.25, 10]\), 
    \(Exercise[\(-.5\), 2.5, 10, laxWendroff, 150, 0.25, 10]\)}], "Input"]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{

Cell["Page 340", "Section"],

Cell[CellGroupData[{

Cell["Problem 8.12", "Subsection"],

Cell[BoxData[
    \(ul = 2; \nur = 0; \nsl = 1; \nu0[x_] := If[x < sl, ul, ur]; \)], "Input"],

Cell[BoxData[
    \(\(uExact[a_, t_]\  := \n\ \ \ \ 
      Plot[u0[x - a\ t], \ {x, \ 2.0, \ 4.0}, \n\t\t
        PlotStyle\  -> \ RGBColor[0, \ 1, \ 0], \n\t\t
        DisplayFunction\  -> \ Identity]; \)\)], "Input"],

Cell["Numerical schemes:", "Text"],

Cell["\<\
To implement the boundary conditions we add an extra element to the \
ends of the array, compute with this element and then remove the two end \
elements.\
\>", "Text"],

Cell[BoxData[
    \(\(upwind[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{vPadded, vNextPadded, vNext}, \n\t\t
        vPadded = Join[{ul}, v1, {ur}]; \n\t\t
        vNextPadded = 
          \((1 - \[Sigma])\)\ vPadded + \[Sigma]\ RotateRight[vPadded]; \n\t\t
        vNext = Take[vNextPadded, {2, Length[vNextPadded] - 1}]; \n\t\t
        Return[{vNext, v1}]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(laxFriedrichs[{v1_List, v0_List}, \ \[Sigma]_]\  := 
      Module[{v1Padded, vNextPadded, vNext}, \n\t\t
        v1Padded = Join[{ul}, v1, {ur}]; \n\t\t
        vNextPadded = 
          \(1\/2\) 
            \((\((1\  + \ \[Sigma])\)\ RotateRight[v1Padded]\  + \ 
                \((1\  - \ \[Sigma])\)\ RotateLeft[v1Padded])\); \n\t\t
        vNext = Take[vNextPadded, {2, Length[vNextPadded] - 1}]; \n\t\t
        Return[{vNext, v1}]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(laxWendroff[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{v1Padded, vNextPadded, vNext}, \n\t\t
        v1Padded = Join[{ul}, v1, {ur}]; \n\t\t
        vNextPadded = 
          \(\[Sigma]\/2\) \((1 + \[Sigma])\) RotateRight[v1Padded] + 
            \((1 - \[Sigma]\^2)\) v1Padded + 
            \(\[Sigma]\/2\) \((\(-1\) + \[Sigma])\) RotateLeft[v1Padded]; \n
        \t\tvNext = Take[vNextPadded, {2, Length[vNextPadded] - 1}]; \n\t\t
        Return[{vNext, v1}]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(leapFrog[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{v0Padded, v1Padded, vNextPadded, v1Next, v0Next}, \n\t\t
        v0Padded = Join[{ul}, v0, {ur}]; \n\t\t
        v1Padded = Join[{ul}, v1, {ur}]; \n\t\t
        vNextPadded = 
          v0Padded - \[Sigma]\ RotateLeft[v1Padded] + 
            \[Sigma]\ RotateRight[v1Padded]; \n\t\t
        v1Next = Take[vNextPadded, {2, Length[vNextPadded] - 1}]; \n\t\t
        v0Next = Take[v1Padded, {2, Length[v1Padded] - 1}]; \n\t\t
        Return[{v1Next, v0Next}]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(Exercise[xL_, xR_, scheme_, \ nCells_, \ \[Sigma]_, \ nTimeSteps_] := 
      \ Module[{a\  = \ 1, \ 
          \[CapitalDelta]x\  = \ \((xR\  - \ xL)\)\/nCells, \ ue, \ au, \ 
          lpa}, \ \[IndentingNewLine]ue\  = \ 
          uExact[a, \[Sigma] \( \[CapitalDelta]x\/a\) nTimeSteps]; \ 
        \[IndentingNewLine]ua\  = \ 
          advectionOneDimension[xL, \ xR, Function[x, u0[x]], \ nCells, \ 
            scheme, \ \[Sigma], \ nTimeSteps]; \ 
        \[IndentingNewLine]lpa\  = \ 
          ListPlot[ua, DisplayFunction\  -> \ Identity]; \n\t\t\n\t\t
        Show[{ue, lpa}, \ \n\ \ \t\t\tPlotRange -> {{2.0, 4.0}, All}, \n
          \ \ \ \ \ \ \ \ \ \ DisplayFunction\  -> \ $DisplayFunction\n
          \t\t\t]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(Exercise[0, 4, upwind, 80, .8, 50]\)], "Input"],

Cell[BoxData[
    \(Exercise[0, 4, laxFriedrichs, 80, .8, 50]\)], "Input"],

Cell[BoxData[
    \(Exercise[0, 4, laxWendroff\ , 80, .8, 50]\)], "Input"],

Cell[BoxData[
    \(Exercise[0, 4, leapFrog\ , 80, .8, 50]\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Problem 8.13", "Subsection"],

Cell[BoxData[
    \(pL = 1.5; \npR = 3.5; \)], "Input"],

Cell[BoxData[
    \(\(u0[x_] := If[0 < x < 1, Sin[2\ \[Pi]\ n\ x], 0]; \)\)], "Input"],

Cell[BoxData[
    \(n = 2; \nPlot[u0[x], {x, \(-2\), 2}]; \nn = 4; \n
    Plot[u0[x], {x, \(-2\), 2}]; \)], "Input"],

Cell[BoxData[
    \(\(uExact[a_, t_]\  := \n\ \ \ \ 
      Plot[u0[x - a\ t], \ {x, \ pL, \ pR}, \n\t\t
        PlotStyle\  -> \ RGBColor[0, \ 1, \ 0], \n\t\t
        DisplayFunction\  -> \ Identity]; \)\)], "Input"],

Cell[BoxData[
    \(\(uExact[1, 2.1] // Show[#, DisplayFunction -> $DisplayFunction]&; 
    \)\)], "Input"],

Cell["Numerical schemes:", "Text"],

Cell[BoxData[
    \(\(upwind[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{}, \n\t\t
        \(Return[{\((1 - \[Sigma])\)\ v1 + \[Sigma]\ RotateRight[v1], v1}]; 
        \)\n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(laxFriedrichs[{v1_List, v0_List}, \ \[Sigma]_]\  := 
      Module[{}, \n\t\t
        \(Return[{
            \(1\/2\) 
              \((\((1\  + \ \[Sigma])\)\ RotateRight[v1]\  + \ 
                  \((1\  - \ \[Sigma])\)\ RotateLeft[v1])\), v1}]; \)\n\t\t]; 
    \)\)], "Input"],

Cell[BoxData[
    \(\(laxWendroff[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{}, \n\t\t
        \(Return[{
            \(\[Sigma]\/2\) \((1 + \[Sigma])\) RotateRight[v1] + 
              \((1 - \[Sigma]\^2)\) v1 + 
              \(\[Sigma]\/2\) \((\(-1\) + \[Sigma])\) RotateLeft[v1], v1}]; 
        \)\n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(leapFrog[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{}, \n\t\t
        \(Return[{v0 - \[Sigma]\ RotateLeft[v1] + \[Sigma]\ RotateRight[v1], 
            v1}]; \)\n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(Exercise[xL_, xR_, scheme_, \ nCells_, \ \[Sigma]_, \ nTimeSteps_] := 
      \ Module[{a\  = \ 1, \ 
          \[CapitalDelta]x\  = \ \((xR\  - \ xL)\)\/nCells, \ ue, \ au, \ 
          lpa}, \ \[IndentingNewLine]ue\  = \ 
          uExact[a, \[Sigma] \( \[CapitalDelta]x\/a\) nTimeSteps]; \ 
        \[IndentingNewLine]ua\  = \ 
          advectionOneDimension[xL, \ xR, Function[x, u0[x]], \ nCells, \ 
            scheme, \ \[Sigma], \ nTimeSteps]; \ 
        \[IndentingNewLine]lpa\  = \ 
          ListPlot[ua, DisplayFunction\  -> \ Identity]; \n\t\t\n\t\t
        Show[{ue, lpa}, \ \n\ \ \t\t\tPlotRange -> {{pL, pR}, All}, \n
          \ \ \ \ \ \ \ \ \ \ DisplayFunction\  -> \ $DisplayFunction\n
          \t\t\t]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[{
    \(n = 2; \nExercise[0, 3.5, upwind, 70, .8, 50]\), 
    \(Exercise[0, 3.5, laxFriedrichs, 70, .8, 50]\), 
    \(Exercise[0, 3.5, laxWendroff\ , 70, .8, 50]\), 
    \(Exercise[0, 3.5, leapFrog\ , 80, .8, 50]\)}], "Input"],

Cell[BoxData[{
    \(n = 4; \nExercise[0, 3.5, upwind, 70, .8, 50]\), 
    \(Exercise[0, 3.5, laxFriedrichs, 70, .8, 50]\), 
    \(Exercise[0, 3.5, laxWendroff\ , 70, .8, 50]\), 
    \(Exercise[0, 3.5, leapFrog\ , 80, .8, 50]\)}], "Input"],

Cell[BoxData[{
    \(n = 2; \nCFL = 0.2; \nExercise[0, 3.5, upwind, 70, CFL, 50]\), 
    \(Exercise[0, 3.5, laxFriedrichs, 70, CFL, 50]\), 
    \(Exercise[0, 3.5, laxWendroff\ , 70, CFL, 50]\), 
    \(Exercise[0, 3.5, leapFrog\ , 80, CFL, 50]\)}], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Problem 8.16", "Subsection"],

Cell[BoxData[
    \(ul = 1; \nur = \(-1\); \nsLoc = 1; \ 
     (*\(+0.025\); \ 
       (*\ Offset\ added\ to\ have\ shock\ location\ NOT\ at\ a\ grid\ 
          \(point.\)*) \n
      \(u0[x_] := If[x < sLoc, ul, ur]; \)\)], "Input"],

Cell[BoxData[
    \(\(uExact[a_, t_]\  := \n\ \ \ \ 
      Plot[u0[x], \ {x, 0\ .0, \ 2.0}, \n\t\t
        PlotStyle\  -> \ RGBColor[0, \ 1, \ 0], \n\t\t
        DisplayFunction\  -> \ Identity]; \)\)], "Input"],

Cell[BoxData[
    \(\(uExact[1, 1] // Show[#, DisplayFunction -> $DisplayFunction]&; \)\)], 
  "Input"],

Cell["Numerical schemes:", "Text"],

Cell["\<\
To implement the boundary conditions we add an extra element to the \
ends of the array, compute with this element and then remove the two end \
elements.\
\>", "Text"],

Cell[BoxData[
    \(\(leapFrog[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{v0Padded, v1Padded, vNextPadded, v1Next, v0Next}, \n\t\t
        v0Padded = Join[{ul}, v0, {ur}]; \n\t\t
        v1Padded = Join[{ul}, v1, {ur}]; \n\t\t
        vNextPadded = 
          v0Padded - \[Sigma]\ v1Padded\ RotateLeft[v1Padded] + 
            \[Sigma]\ v1Padded\ RotateRight[v1Padded]; \n\t\t
        v1Next = Take[vNextPadded, {2, Length[vNextPadded] - 1}]; \n\t\t
        v0Next = Take[v1Padded, {2, Length[v1Padded] - 1}]; \n\t\t
        Return[{v1Next, v0Next}]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(Exercise[xL_, xR_, scheme_, \ nCells_, \ \[Sigma]_, \ nTimeSteps_] := 
      \ Module[{a\  = \ 1, \ 
          \[CapitalDelta]x\  = \ \((xR\  - \ xL)\)\/nCells, \ ue, \ au, \ 
          lpa}, \ \[IndentingNewLine]ue\  = \ 
          uExact[a, \[Sigma] \( \[CapitalDelta]x\/a\) nTimeSteps]; \ 
        \[IndentingNewLine]ua\  = \ 
          advectionOneDimension[xL, \ xR, Function[x, u0[x]], \ nCells, \ 
            scheme, \ \[Sigma], \ nTimeSteps]; \ 
        \[IndentingNewLine]lpa\  = \ 
          ListPlot[ua, DisplayFunction\  -> \ Identity]; \n\t\t\n\t\t
        Show[{ue, lpa}, \ \n\ \ \t\t\tPlotRange -> {{0.0, 2.0}, All}, \n
          \ \ \ \ \ \ \ \ \ \ DisplayFunction\  -> \ $DisplayFunction\n
          \t\t\t]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[{
    \(CFL = 0.8; \nExercise[0, 2, leapFrog\ , 40, CFL, 0]\), 
    \(Exercise[0, 2, leapFrog\ , 40, CFL, 1]\), 
    \(Exercise[0, 2, leapFrog\ , 40, CFL, 2]\)}], "Input"]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{

Cell["Page 341", "Section"],

Cell[CellGroupData[{

Cell["Problem 8.21", "Subsection"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Simplify[
      \((\(-Sin[\[Phi]]\) - \(3\/2\) I\ \[Sigma]\ Cos[\[Phi]])\) 
          \((2 + Cos[\[Phi]] + \(3\/2\) I\ \[Sigma]\ Sin[\[Phi]])\) - 
        \((2 + Cos[\[Phi]] - \(3\/2\) I\ \[Sigma]\ Sin[\[Phi]])\) 
          \((\(-Sin[\[Phi]]\) + \(3\/2\) I\ \[Sigma]\ Cos[\[Phi]])\)]\)], 
  "Input"],

Cell[BoxData[
    \(\(-3\)\ I\ \[Sigma]\ \((1 + 2\ Cos[\[Phi]])\)\)], "Output"]
}, Open  ]]
}, Closed]],

Cell[CellGroupData[{

Cell["Problem 8.22", "Subsection"],

Cell[BoxData[
    \(<< Graphics`\)], "Input"],

Cell[BoxData[
    \(upwind[\[Sigma]_] := 
      PolarPlot[
        \((\((1 - \[Sigma])\)\ Cos[\[Phi]] + \[Sigma])
            \)\/\(\((1 + \[Sigma] \((Cos[\[Phi]] - 1)\))\)\^2 + 
            \(\[Sigma]\^2\) Sin[\[Phi]]\^2\), {\[Phi], 0, 2  \[Pi]}]\)], 
  "Input"],

Cell[BoxData[
    \(upwind[0.2]; \nupwind[0.4]; \nupwind[0.6]; \nupwind[0.8]; \n
    upwind[1.0]; \)], "Input"],

Cell[BoxData[
    \(laxFriedrichs[\[Sigma]_] := 
      PolarPlot[
        1\/\(Cos[\[Phi]]\^2 + \(\[Sigma]\^2\) Sin[\[Phi]]\^2\), {\[Phi], 0, 
          2  \[Pi]}]\)], "Input"],

Cell[BoxData[
    \(laxFriedrichs[0.2]; \nlaxFriedrichs[0.4]; \nlaxFriedrichs[0.6]; \n
    laxFriedrichs[0.8]; \nlaxFriedrichs[1.0]; \nlaxFriedrichs[1.5]; \)], 
  "Input"],

Cell[BoxData[
    \(laxWendroff[\[Sigma]_] := 
      PolarPlot[
        \(\((1 - 2\ \(\[Sigma]\^2\) Sin[\[Phi]\/2]\^2)\) Cos[\[Phi]] + 
            \(\[Sigma]\^2\) 
              Sin[\[Phi]]\^2\)\/\(\((1 - 2 \( \[Sigma]\^2\) Sin[\[Phi]\/2]\^2)
                \)\^2 + \(\[Sigma]\^2\) Sin[\[Phi]]\^2\), {\[Phi], 0, 
          2  \[Pi]}]\)], "Input"],

Cell[BoxData[
    \(CFL = {0.2, 0.4, 0.6, 0.8, 1.0, 1.2}; \n
    Table[laxWendroff[CFL[\([i]\)]], {i, 1, Length[CFL]}]; \)], "Input"]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{

Cell["Page 353", "Section"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Series[
        ArcTan[\ \[Sigma]\ Tan[\[Phi]]]\/\(\[Sigma]\ \[Phi]\), {\[Phi], 0, 3}]
         // Simplify\)], "Input"],

Cell[BoxData[
    InterpretationBox[
      RowBox[{"1", "+", \(\((1\/3 - \[Sigma]\^2\/3)\)\ \[Phi]\^2\), "+", 
        InterpretationBox[\(O[\[Phi]]\^4\),
          SeriesData[ \[Phi], 0, {}, 0, 4, 1]]}],
      SeriesData[ \[Phi], 0, {1, 0, 
        Plus[ 
          Rational[ 1, 3], 
          Times[ 
            Rational[ -1, 3], 
            Power[ \[Sigma], 2]]]}, 0, 4, 1]]], "Output"]
}, Open  ]]
}, Closed]],

Cell[CellGroupData[{

Cell["Page 355", "Section"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Simplify[1 - 2\ Sin[\[Phi]\/2]\^2 + 8\ Sin[\[Phi]\/2]\^4]\)], "Input"],

Cell[BoxData[
    \(3 - 3\ Cos[\[Phi]] + Cos[2\ \[Phi]]\)], "Output"]
}, Open  ]],

Cell[BoxData[
    \(\(p[n_] := 
      Plot[{Cos[n\ \[Phi]], 
          1 + Sum[\(\((\(-1\))\)\^l\) \(2\^\(2\ l - 1\)\) 
                Sin[\[Phi]\/2]\^\(2\ l\), {l, n}]}, {\[Phi], 0, 2\ \[Pi]}, \n
        \t\tPlotRange -> All]; \)\)], "Input"],

Cell[BoxData[
    \(\(p[3]; \)\)], "Input"],

Cell["\<\
I can't seem to get these two expressions to match?  Could this \
expression be incorrect?  Gradshteyn and Ryzhik has a different version?\
\>",
   "Text"]
}, Closed]],

Cell[CellGroupData[{

Cell["Page 358", "Section"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Solve[{b\_\(-2\) + b\_\(-1\) + b\_0 == 1, \n\t\t
          \(-2\)\ b\_\(-2\) - b\_\(-1\) == \(-\[Sigma]\), \n\t\t
          4  b\_\(-2\) + b\_\(-1\) == \[Sigma]\^2}, \n
        \t{b\_\(-2\), b\_\(-1\), b\_0}] // Simplify\)], "Input"],

Cell[BoxData[
    \({{b\_\(-2\) \[Rule] 1\/2\ \((\(-1\) + \[Sigma])\)\ \[Sigma], 
        b\_\(-1\) \[Rule] \(-\((\(-2\) + \[Sigma])\)\)\ \[Sigma], 
        b\_0 \[Rule] 1\/2\ \((2 - 3\ \[Sigma] + \[Sigma]\^2)\)}}\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Simplify[
      \(\(-8\)\/2\) \[Sigma] \((\[Sigma] - 1)\) - 
        \[Sigma] \((2 - \[Sigma])\) + \[Sigma]\^3]\)], "Input"],

Cell[BoxData[
    \(\[Sigma]\ \((2 - 3\ \[Sigma] + \[Sigma]\^2)\)\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Factor[Out[9]]\)], "Input"],

Cell[BoxData[
    \(\((\(-2\) + \[Sigma])\)\ \((\(-1\) + \[Sigma])\)\ \[Sigma]\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Simplify[
      \(16\/2\) \[Sigma] \((\[Sigma] - 1)\) + \[Sigma] \((2 - \[Sigma])\) - 
        \[Sigma]\^4]\)], "Input"],

Cell[BoxData[
    \(\(-\[Sigma]\)\ \((6 - 7\ \[Sigma] + \[Sigma]\^3)\)\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Factor[Out[11]]\)], "Input"],

Cell[BoxData[
    \(\(-\((\(-2\) + \[Sigma])\)\)\ \((\(-1\) + \[Sigma])\)\ \[Sigma]\ \((
      3 + \[Sigma])\)\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(TrigExpand[
      1 - \[Sigma] \((1 - Cos[\[Phi]] + I\ Sin[\[Phi]])\) + 
        \(1\/2\) \[Sigma] \((\[Sigma] - 1)\) 
          \((1 - 2 \((Cos[\[Phi]] - I\ Sin[\[Phi]])\) + 
              \((Cos[2\ \[Phi]] - I\ Sin[2\ \[Phi]])\))\)]\)], "Input"],

Cell[BoxData[
    \(1 - \(3\ \[Sigma]\)\/2 + \[Sigma]\^2\/2 + 2\ \[Sigma]\ Cos[\[Phi]] - 
      \[Sigma]\^2\ Cos[\[Phi]] - 1\/2\ \[Sigma]\ Cos[\[Phi]]\^2 + 
      1\/2\ \[Sigma]\^2\ Cos[\[Phi]]\^2 - 2\ I\ \[Sigma]\ Sin[\[Phi]] + 
      I\ \[Sigma]\^2\ Sin[\[Phi]] + I\ \[Sigma]\ Cos[\[Phi]]\ Sin[\[Phi]] - 
      I\ \[Sigma]\^2\ Cos[\[Phi]]\ Sin[\[Phi]] + 
      1\/2\ \[Sigma]\ Sin[\[Phi]]\^2 - 1\/2\ \[Sigma]\^2\ Sin[\[Phi]]\^2\)], 
  "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Simplify[
      \(-\(\(3\ \[Sigma]\)\/2\)\) + \[Sigma]\^2\/2 + 
        2\ \[Sigma]\ Cos[\[Phi]] - \[Sigma]\^2\ Cos[\[Phi]] - 
        1\/2\ \[Sigma]\ Cos[\[Phi]]\^2 + 1\/2\ \[Sigma]\^2\ Cos[\[Phi]]\^2 + 
        1\/2\ \[Sigma]\ Sin[\[Phi]]\^2 - 1\/2\ \[Sigma]\^2\ Sin[\[Phi]]\^2]
      \)], "Input"],

Cell[BoxData[
    \(\(-2\)\ \[Sigma]\ \((1 + \((\(-1\) + \[Sigma])\)\ Cos[\[Phi]])\)\ 
      Sin[\[Phi]\/2]\^2\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Simplify[
      \(-2\)\ I\ \[Sigma]\ Sin[\[Phi]] + I\ \[Sigma]\^2\ Sin[\[Phi]] + 
        I\ \[Sigma]\ Cos[\[Phi]]\ Sin[\[Phi]] - 
        I\ \[Sigma]\^2\ Cos[\[Phi]]\ Sin[\[Phi]]]\)], "Input"],

Cell[BoxData[
    \(\(-I\)\ \[Sigma]\ \((
      2 - \[Sigma] + \((\(-1\) + \[Sigma])\)\ Cos[\[Phi]])\)\ Sin[\[Phi]]\)], 
  "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Simplify[
      \((1 - 2\ \[Sigma]\ \((1 + \((\(-1\) + \[Sigma])\)\ Cos[\[Phi]])\)\ 
                Sin[\[Phi]\/2]\^2)\)\^2 + 
        \((\ \[Sigma]\ \((2 - \[Sigma] + \((\(-1\) + \[Sigma])\)\ Cos[\[Phi]])
              \)\ Sin[\[Phi]])\)\^2]\)], "Input"],

Cell[BoxData[
    \(\((1 - 
            2\ \[Sigma]\ \((1 + \((\(-1\) + \[Sigma])\)\ Cos[\[Phi]])\)\ 
              Sin[\[Phi]\/2]\^2)\)\^2 + 
      \[Sigma]\^2\ 
        \((2 - \[Sigma] + \((\(-1\) + \[Sigma])\)\ Cos[\[Phi]])\)\^2\ 
        Sin[\[Phi]]\^2\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(TrigExpand[Out[31]]\)], "Input"],

Cell[BoxData[
    \(1 - 3\ \[Sigma] + \(15\ \[Sigma]\^2\)\/2 - 6\ \[Sigma]\^3 + 
      \(3\ \[Sigma]\^4\)\/2 + 4\ \[Sigma]\ Cos[\[Phi]\/2]\^2 - 
      10\ \[Sigma]\^2\ Cos[\[Phi]\/2]\^2 + 
      8\ \[Sigma]\^3\ Cos[\[Phi]\/2]\^2 - 2\ \[Sigma]\^4\ Cos[\[Phi]\/2]\^2 - 
      \[Sigma]\ Cos[\[Phi]\/2]\^4 + 5\/2\ \[Sigma]\^2\ Cos[\[Phi]\/2]\^4 - 
      2\ \[Sigma]\^3\ Cos[\[Phi]\/2]\^4 + 
      1\/2\ \[Sigma]\^4\ Cos[\[Phi]\/2]\^4 - 4\ \[Sigma]\ Sin[\[Phi]\/2]\^2 + 
      10\ \[Sigma]\^2\ Sin[\[Phi]\/2]\^2 - 
      8\ \[Sigma]\^3\ Sin[\[Phi]\/2]\^2 + 2\ \[Sigma]\^4\ Sin[\[Phi]\/2]\^2 + 
      6\ \[Sigma]\ Cos[\[Phi]\/2]\^2\ Sin[\[Phi]\/2]\^2 - 
      15\ \[Sigma]\^2\ Cos[\[Phi]\/2]\^2\ Sin[\[Phi]\/2]\^2 + 
      12\ \[Sigma]\^3\ Cos[\[Phi]\/2]\^2\ Sin[\[Phi]\/2]\^2 - 
      3\ \[Sigma]\^4\ Cos[\[Phi]\/2]\^2\ Sin[\[Phi]\/2]\^2 - 
      \[Sigma]\ Sin[\[Phi]\/2]\^4 + 5\/2\ \[Sigma]\^2\ Sin[\[Phi]\/2]\^4 - 
      2\ \[Sigma]\^3\ Sin[\[Phi]\/2]\^4 + 
      1\/2\ \[Sigma]\^4\ Sin[\[Phi]\/2]\^4\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Factor[
      \(-3\)\ \[Sigma] + \(15\ \[Sigma]\^2\)\/2 - 6\ \[Sigma]\^3 + 
        \(3\ \[Sigma]\^4\)\/2]\)], "Input"],

Cell[BoxData[
    \(3\/2\ \((\(-2\) + \[Sigma])\)\ \((\(-1\) + \[Sigma])\)\^2\ \[Sigma]\)], 
  "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Factor[
      \(+4\)\ \[Sigma]\ Cos[\[Phi]\/2]\^2 - 
        10\ \[Sigma]\^2\ Cos[\[Phi]\/2]\^2 + 
        8\ \[Sigma]\^3\ Cos[\[Phi]\/2]\^2 - 2\ \[Sigma]\^4\ Cos[\[Phi]\/2]\^2]
      \)], "Input"],

Cell[BoxData[
    \(\(-2\)\ \((\(-2\) + \[Sigma])\)\ \((\(-1\) + \[Sigma])\)\^2\ \[Sigma]\ 
      Cos[\[Phi]\/2]\^2\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Factor[
      \(-\[Sigma]\)\ Cos[\[Phi]\/2]\^4 + 
        5\/2\ \[Sigma]\^2\ Cos[\[Phi]\/2]\^4 - 
        2\ \[Sigma]\^3\ Cos[\[Phi]\/2]\^4 + 
        1\/2\ \[Sigma]\^4\ Cos[\[Phi]\/2]\^4]\)], "Input"],

Cell[BoxData[
    \(1\/2\ \((\(-2\) + \[Sigma])\)\ \((\(-1\) + \[Sigma])\)\^2\ \[Sigma]\ 
      Cos[\[Phi]\/2]\^4\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Factor[
      \(-4\)\ \[Sigma]\ Sin[\[Phi]\/2]\^2 + 
        10\ \[Sigma]\^2\ Sin[\[Phi]\/2]\^2 - 
        8\ \[Sigma]\^3\ Sin[\[Phi]\/2]\^2 + 2\ \[Sigma]\^4\ Sin[\[Phi]\/2]\^2]
      \)], "Input"],

Cell[BoxData[
    \(2\ \((\(-2\) + \[Sigma])\)\ \((\(-1\) + \[Sigma])\)\^2\ \[Sigma]\ 
      Sin[\[Phi]\/2]\^2\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Factor[
      6\ \[Sigma]\ Cos[\[Phi]\/2]\^2\ Sin[\[Phi]\/2]\^2 - 
        15\ \[Sigma]\^2\ Cos[\[Phi]\/2]\^2\ Sin[\[Phi]\/2]\^2 + 
        12\ \[Sigma]\^3\ Cos[\[Phi]\/2]\^2\ Sin[\[Phi]\/2]\^2 - 
        3\ \[Sigma]\^4\ Cos[\[Phi]\/2]\^2\ Sin[\[Phi]\/2]\^2]\)], "Input"],

Cell[BoxData[
    \(\(-3\)\ \((\(-2\) + \[Sigma])\)\ \((\(-1\) + \[Sigma])\)\^2\ \[Sigma]\ 
      Cos[\[Phi]\/2]\^2\ Sin[\[Phi]\/2]\^2\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Factor[
      \(-\[Sigma]\)\ Sin[\[Phi]\/2]\^4 + 
        5\/2\ \[Sigma]\^2\ Sin[\[Phi]\/2]\^4 - 
        2\ \[Sigma]\^3\ Sin[\[Phi]\/2]\^4 + 
        1\/2\ \[Sigma]\^4\ Sin[\[Phi]\/2]\^4]\)], "Input"],

Cell[BoxData[
    \(1\/2\ \((\(-2\) + \[Sigma])\)\ \((\(-1\) + \[Sigma])\)\^2\ \[Sigma]\ 
      Sin[\[Phi]\/2]\^4\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Expand[
      3\/2 - 2 \((1 - x\^2)\) + \(1\/2\) \((1 - x\^2)\)\^2 + 2\ x\^2 - 
        3\ \(x\^2\) \((1 - x\^2)\) + \(1\/2\) x\^4]\)], "Input"],

Cell[BoxData[
    \(4\ x\^4\)], "Output"]
}, Open  ]],

Cell[BoxData[
    \(\(Plot[{x \(\((1 - x)\)\^2\) \((2 - x)\), 1\/2}, {x, 0, 4}]; \)\)], 
  "Input"],

Cell[BoxData[
    \(\(Plot[1\/Sin[x]\^4, {x, 1, \[Pi]\/2}]; \)\)], "Input"],

Cell[BoxData[
    \(\(Plot[\(1\/6\) \((1 - x)\) \((2 - x)\), {x, 0, 2}]; \)\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Page 367", "Section"],

Cell["Problem 9.1", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Series[\ 
        ArcSin[\ \[Sigma]\ Sin[\[Phi]]]\/\(\[Sigma]\ \[Phi]\), {\[Phi], 0, 6}]
         // Simplify\)], "Input"],

Cell[BoxData[
    InterpretationBox[
      RowBox[{
      "1", "+", \(1\/6\ \((\(-1\) + \[Sigma]\^2)\)\ \[Phi]\^2\), "+", 
        \(1\/120\ \((1 - 10\ \[Sigma]\^2 + 9\ \[Sigma]\^4)\)\ \[Phi]\^4\), 
        "+", \(\(\((
                \(-1\) + 91\ \[Sigma]\^2 - 315\ \[Sigma]\^4 + 
                  225\ \[Sigma]\^6)\)\ \[Phi]\^6\)\/5040\), "+", 
        InterpretationBox[\(O[\[Phi]]\^7\),
          SeriesData[ \[Phi], 0, {}, 0, 7, 1]]}],
      SeriesData[ \[Phi], 0, {1, 0, 
        Times[ 
          Rational[ 1, 6], 
          Plus[ -1, 
            Power[ \[Sigma], 2]]], 0, 
        Times[ 
          Rational[ 1, 120], 
          Plus[ 1, 
            Times[ -10, 
              Power[ \[Sigma], 2]], 
            Times[ 9, 
              Power[ \[Sigma], 4]]]], 0, 
        Times[ 
          Rational[ 1, 5040], 
          Plus[ -1, 
            Times[ 91, 
              Power[ \[Sigma], 2]], 
            Times[ -315, 
              Power[ \[Sigma], 4]], 
            Times[ 225, 
              Power[ \[Sigma], 6]]]]}, 0, 7, 1]]], "Output"]
}, Open  ]],

Cell["Problem 9.2", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Series[
        ArcTan[\((\[Sigma]\ Sin[\[Phi]])\)\/\(1 - \[Sigma] - 
                \[Sigma]\ Cos[\[Phi]]\)]\/\(\[Sigma]\ \[Phi]\), {\[Phi], 0, 
          4}] // Simplify\)], "Input"],

Cell[BoxData[
    InterpretationBox[
      RowBox[{
      \(1\/\(1 - 2\ \[Sigma]\)\), "-", 
        \(\(\((\(-1\) + \[Sigma])\)\ \[Phi]\^2\)\/\(6\ 
              \((\(-1\) + 2\ \[Sigma])\)\^3\)\), "-", 
        \(\(\((1 + 7\ \[Sigma] - 16\ \[Sigma]\^2 + 8\ \[Sigma]\^3)\)\ 
              \[Phi]\^4\)\/\(120\ \((\(-1\) + 2\ \[Sigma])\)\^5\)\), "+", 
        InterpretationBox[\(O[\[Phi]]\^5\),
          SeriesData[ \[Phi], 0, {}, 0, 5, 1]]}],
      SeriesData[ \[Phi], 0, {
        Power[ 
          Plus[ 1, 
            Times[ -2, \[Sigma]]], -1], 0, 
        Times[ 
          Rational[ -1, 6], 
          Plus[ -1, \[Sigma]], 
          Power[ 
            Plus[ -1, 
              Times[ 2, \[Sigma]]], -3]], 0, 
        Times[ 
          Rational[ -1, 120], 
          Power[ 
            Plus[ -1, 
              Times[ 2, \[Sigma]]], -5], 
          Plus[ 1, 
            Times[ 7, \[Sigma]], 
            Times[ -16, 
              Power[ \[Sigma], 2]], 
            Times[ 8, 
              Power[ \[Sigma], 3]]]]}, 0, 5, 1]]], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Series[\@\(Cos[\[Phi]]\^2 + \[Sigma]\ Sin[\[Phi]]\^2\), {\[Phi], 0, 3}]
      \)], "Input"],

Cell[BoxData[
    InterpretationBox[
      RowBox[{"1", "+", \(1\/2\ \((\(-1\) + \[Sigma])\)\ \[Phi]\^2\), "+", 
        InterpretationBox[\(O[\[Phi]]\^4\),
          SeriesData[ \[Phi], 0, {}, 0, 4, 1]]}],
      SeriesData[ \[Phi], 0, {1, 0, 
        Times[ 
          Rational[ 1, 2], 
          Plus[ -1, \[Sigma]]]}, 0, 4, 1]]], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Series[
        ArcTan[\[Sigma]\ Tan[\[Phi]]]\/\(\[Sigma]\ \[Phi]\), {\[Phi], 0, 3}]
         // Simplify\)], "Input"],

Cell[BoxData[
    InterpretationBox[
      RowBox[{"1", "+", \(\((1\/3 - \[Sigma]\^2\/3)\)\ \[Phi]\^2\), "+", 
        InterpretationBox[\(O[\[Phi]]\^4\),
          SeriesData[ \[Phi], 0, {}, 0, 4, 1]]}],
      SeriesData[ \[Phi], 0, {1, 0, 
        Plus[ 
          Rational[ 1, 3], 
          Times[ 
            Rational[ -1, 3], 
            Power[ \[Sigma], 2]]]}, 0, 4, 1]]], "Output"]
}, Open  ]]
}, Closed]],

Cell[CellGroupData[{

Cell["Page 368", "Section"],

Cell[CellGroupData[{

Cell["Problem 9.7", "Subsection"],

Cell[BoxData[
    \(\(Needs["\<Graphics`Graphics`\>"]; \)\)], "Input"],

Cell[BoxData[
    \(\(GSquared[\[Phi]_, \[Sigma]_] := 
      \((1 - 2  \[Sigma] \((1 - \((1 - \[Sigma])\) Cos[\[Phi]])\)\ 
                Sin[\[Phi]\/2]\^2)\)\^2 + 
        \(\[Sigma]\^2\) \(Sin[\[Phi]]\^2\) 
          \((1 + 2 \((1 - \[Sigma])\) Sin[\[Phi]\/2]\^2)\)\^2; \)\)], "Input"],

Cell[BoxData[
    \(\(pp[\[Sigma]_] := 
      PolarPlot[{1, GSquared[\[Phi], \[Sigma]]}, {\[Phi], 0, 2  \[Pi]}]; 
    \)\)], "Input"],

Cell[BoxData[
    \(samples = {0.25, 0.5, 0.75, 0.99, 1.1}; \n
    For[i = 1, i <= Length[samples], \(i++\), \n\t\(pp[samples[\([i]\)]]; \)\n
      \t]; \)], "Input"],

Cell["Which of these two formulations is correct?", "Text"],

Cell[BoxData[
    \(\(\[CapitalPhi][\[Sigma]_, \[Phi]_] := 
      ArcTan[\[Sigma]\ Sin[\[Phi]] 
          \((1 + 2 \((1 - \[Sigma])\) Sin[\[Phi]\/2]\^2)\), 
        1 - 2  \[Sigma] \((1 - \((1 - \[Sigma])\) Cos[\[Sigma]])\) 
            Sin[\[Phi]\/2]\^2]; \)\)], "Input"],

Cell[BoxData[
    \(\(\[CapitalPhi][\[Sigma]_, \[Phi]_] := 
      ArcTan[\(\[Sigma]\ Sin[\[Phi]] 
            \((1 + 2 \((1 - \[Sigma])\) Sin[\[Phi]\/2]\^2)\)\)\/\(1 - 
            2  \[Sigma] \((1 - \((1 - \[Sigma])\) Cos[\[Sigma]])\) 
              Sin[\[Phi]\/2]\^2\)]; \)\)], "Input"],

Cell[BoxData[
    \(\(\[Epsilon]\_\[Phi][\[Sigma]_, \[Phi]_] := 
      \[CapitalPhi][\[Sigma], \[Phi]]\/\(\[Sigma]\ \[Phi]\); \)\)], "Input"],

Cell[BoxData[
    \(\(pp[\[Sigma]_] := 
      PolarPlot[{1, \[Epsilon]\_\[Phi][\[Sigma], \[Phi]]}, {\[Phi], 0, 
          2\ \[Pi]}]; \)\)], "Input"],

Cell[BoxData[
    \(samples = {0.25, 0.5, 0.75, 0.99, 1.1}; \n
    For[i = 1, i <= Length[samples], \(i++\), \n\t\(pp[samples[\([i]\)]]; \)\n
      \t]; \)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Problem 9.8", "Subsection"],

Cell["\<\
Exact solution.  The definitions of ul and ur are needed for the \
boundary condition routines.\
\>", "Text"],

Cell[BoxData[
    \(ul = 0; \nur = 0; \n
    u0[x_, k_] := If[0 < x < 1, Sin[2\ \[Pi]\ k\ x], 0]; \)], "Input"],

Cell[BoxData[
    \(\(uExact[a_, k_, t_]\  := \n\ \ \ \ 
      Plot[u0[x - a\ t, k], \ {x, \ \(-1.0\), \ 6.0}, \n\t\t
        PlotStyle\  -> \ RGBColor[0, \ 1, \ 0], \n\t\t
        DisplayFunction\  -> \ Identity]; \)\)], "Input"],

Cell[BoxData[
    \(\(Show[uExact[1, 4, 1], DisplayFunction -> $DisplayFunction]; \)\)], 
  "Input"],

Cell["Numerical schemes:", "Text"],

Cell["\<\
To implement the periodic boundary conditions for this problem we \
add an extra element to the ends of the array, compute with this element and \
then remove the two end elements.  The vector of unknowns values at time \
\"n\" are stored in v1 with the previous values at time \"n-1\" stored in v0.\
\
\>", "Text"],

Cell[BoxData[
    \(\(upwind[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{vPadded, vNextPadded, vNext}, \n\t\t
        vPadded = Join[{ul}, v1, {ur}]; \n\t\t
        vNextPadded = 
          \((1 - \[Sigma])\)\ vPadded + \[Sigma]\ RotateRight[vPadded]; \n\t\t
        vNext = Take[vNextPadded, {2, Length[vNextPadded] - 1}]; \n\t\t
        Return[{vNext, v1}]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(laxFriedrichs[{v1_List, v0_List}, \ \[Sigma]_]\  := 
      Module[{v1Padded, vNextPadded, vNext}, \n\t\t
        v1Padded = Join[{ul}, v1, {ur}]; \n\t\t
        vNextPadded = 
          \(1\/2\) 
            \((\((1\  + \ \[Sigma])\)\ RotateRight[v1Padded]\  + \ 
                \((1\  - \ \[Sigma])\)\ RotateLeft[v1Padded])\); \n\t\t
        vNext = Take[vNextPadded, {2, Length[vNextPadded] - 1}]; \n\t\t
        Return[{vNext, v1}]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(laxWendroff[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{v1Padded, vNextPadded, vNext}, \n\t\t
        v1Padded = Join[{ul}, v1, {ur}]; \n\t\t
        vNextPadded = 
          \(\[Sigma]\/2\) \((1 + \[Sigma])\) RotateRight[v1Padded] + 
            \((1 - \[Sigma]\^2)\) v1Padded + 
            \(\[Sigma]\/2\) \((\(-1\) + \[Sigma])\) RotateLeft[v1Padded]; \n
        \t\tvNext = Take[vNextPadded, {2, Length[vNextPadded] - 1}]; \n\t\t
        Return[{vNext, v1}]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(leapFrog[{v1_List, v0_List}, \[Sigma]_] := 
      Module[{v0Padded, v1Padded, vNextPadded, v1Next, v0Next}, \n\t\t
        v0Padded = Join[{ul}, v0, {ur}]; \n\t\t
        v1Padded = Join[{ul}, v1, {ur}]; \n\t\t
        vNextPadded = 
          v0Padded - \[Sigma]\ RotateLeft[v1Padded] + 
            \[Sigma]\ RotateRight[v1Padded]; \n\t\t
        v1Next = Take[vNextPadded, {2, Length[vNextPadded] - 1}]; \n\t\t
        v0Next = Take[v1Padded, {2, Length[v1Padded] - 1}]; \n\t\t
        Return[{v1Next, v0Next}]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(\(Exercise[xL_, xR_, scheme_, \[CapitalDelta]x_, \ \[Sigma]_, \ 
        nTimeSteps_, k_] := \ 
      Module[{a\  = \ 1, \ nCells\  = \ \((xR\  - \ xL)\)\/\[CapitalDelta]x, 
          \ ue, \ ua, \ lpa}, \ 
        \[IndentingNewLine]ue\  = \ 
          uExact[a, k, \[Sigma] \( \[CapitalDelta]x\/a\) nTimeSteps]; \ 
        \[IndentingNewLine]ua\  = \ 
          advectionOneDimension[xL, \ xR, Function[x, u0[x, k]], \ nCells, \ 
            scheme, \ \[Sigma], \ nTimeSteps]; \ 
        \[IndentingNewLine]lpa\  = \ 
          ListPlot[ua, DisplayFunction\  -> \ Identity]; \n\t\t\n\t\t
        Show[{ue, lpa}, \ \n\ \ \t\t\tPlotRange -> Automatic, \n
          \ \ \ \ \ \ \ \ \ \ DisplayFunction\  -> \ $DisplayFunction\n
          \t\t\t]; \n\t\t]; \)\)], "Input"],

Cell[BoxData[
    \(xL = \(-1\); xR = 3; \n\[CapitalDelta]x = 1\/60; \nk = 4; \)], "Input"],

Cell[BoxData[{
    \(scheme = upwind; \n\n\[Sigma] = 0.1; \n
    Exercise[xL, xR, scheme, \[CapitalDelta]x, \[Sigma], 50, k]\), 
    \(Exercise[xL, xR, scheme, \[CapitalDelta]x, \[Sigma], 100, k]\), 
    \(Exercise[xL, xR, scheme, \[CapitalDelta]x, \[Sigma], 200, k]\n\), 
    \(\[Sigma] = 0.5; \n
    Exercise[xL, xR, scheme, \[CapitalDelta]x, \[Sigma], 50, k]\), 
    \(Exercise[xL, xR, scheme, \[CapitalDelta]x, \[Sigma], 100, k]\), 
    \(Exercise[xL, xR, scheme, \[CapitalDelta]x, \[Sigma], 200, k]\n\), 
    \(\[Sigma] = 0.9; \n
    Exercise[xL, xR, scheme, \[CapitalDelta]x, \[Sigma], 50, k]\), 
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Cell["Problem 9.9", "Subsection"],

Cell["\<\
Exact solution.  The definitions of ul and ur are needed for the \
boundary condition routines.\
\>", "Text"],

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Cell[BoxData[
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      Plot[u0[x - a\ t], \ {x, \ xL, xR}, \n\t\t
        PlotStyle\  -> \ RGBColor[0, \ 1, \ 0], \n\t\t
        DisplayFunction\  -> \ Identity]; \)\)], "Input"],

Cell[BoxData[
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Cell["Numerical schemes:", "Text"],

Cell["\<\
To implement the periodic boundary conditions for this problem we \
add an extra element to the ends of the array, compute with this element and \
then remove the two end elements.  The vector of unknowns values at time \
\"n\" are stored in v1 with the previous values at time \"n-1\" stored in v0.\
\
\>", "Text"],

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        vNextPadded = 
          \((1 - \[Sigma])\)\ vPadded + \[Sigma]\ RotateRight[vPadded]; \n\t\t
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        vNextPadded = 
          \(1\/2\) 
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                \((1\  - \ \[Sigma])\)\ RotateLeft[v1Padded])\); \n\t\t
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Cell["\<\
Note: the second order upwind scheme is also known as the method of \
Beam and Warming\
\>", "Text"],

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          \ ue, \ ua, \ lpa}, \ 
        \[IndentingNewLine]ue\  = \ 
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        \[IndentingNewLine]ua\  = \ 
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        \[IndentingNewLine]lpa\  = \ 
          ListPlot[ua, DisplayFunction\  -> \ Identity]; \n\t\t\n\t\t
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*)




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End of Mathematica Notebook file.
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